If \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C, where 0 < x < \(\frac{\pi}{4}\) , then which one of the following is correct?
A + B - 2 = 0
Solving the Integral \(\int \sqrt{1 - \sin 2x} \, dx\) for \(0 < x < \frac{\pi}{4}\)
The problem asks us to evaluate the integral \(\int \sqrt{1 - \sin 2x} \, dx\) and express the result in the form \(A \sin x + B \cos x + C\), given that \(0 < x < \frac{\pi}{4}\). We then need to find the correct relationship between the coefficients A and B.
Simplifying the Integrand: \(\sqrt{1 - \sin 2x}\)
First, let's simplify the expression under the square root. We know the trigonometric identities:
- \(1 = \sin^2 x + \cos^2 x\)
- \(\sin 2x = 2 \sin x \cos x\)
Using these identities, we can rewrite \(1 - \sin 2x\) as:
\(1 - \sin 2x = (\sin^2 x + \cos^2 x) - 2 \sin x \cos x\)
This expression is in the form of a perfect square, \((a - b)^2 = a^2 - 2ab + b^2\) or \((b - a)^2 = b^2 - 2ab + a^2\). Specifically, it matches \((\cos x - \sin x)^2\) or \((\sin x - \cos x)^2\).
\(1 - \sin 2x = (\cos x - \sin x)^2\) or \((\sin x - \cos x)^2\)
So, \(\sqrt{1 - \sin 2x} = \sqrt{(\cos x - \sin x)^2}\). The square root of a square is the absolute value: \(\sqrt{y^2} = |y|\).
Therefore, \(\sqrt{1 - \sin 2x} = |\cos x - \sin x|\).
Considering the Given Range: \(0 < x < \frac{\pi}{4}\)
The problem specifies the range \(0 < x < \frac{\pi}{4}\). In this interval:
- The value of \(\cos x\) decreases from 1 towards \(\frac{1}{\sqrt{2}}\).
- The value of \(\sin x\) increases from 0 towards \(\frac{1}{\sqrt{2}}\).
For any \(x\) strictly between 0 and \(\frac{\pi}{4}\), \(\cos x\) is greater than \(\sin x\).
This means that \(\cos x - \sin x\) is positive in this range.
Hence, for \(0 < x < \frac{\pi}{4}\), \(|\cos x - \sin x| = \cos x - \sin x\).
So, the integrand simplifies to \(\cos x - \sin x\).
Performing the Integration
Now we integrate the simplified expression:
\(\int \sqrt{1 - \sin 2x} \, dx = \int (\cos x - \sin x) \, dx\)
We can integrate term by term:
\(\int \cos x \, dx = \sin x + C_1\)
\(\int \sin x \, dx = -\cos x + C_2\)
Combining these, we get:
\(\int (\cos x - \sin x) \, dx = \int \cos x \, dx - \int \sin x \, dx\)
\(= (\sin x) - (-\cos x) + C\)
\(= \sin x + \cos x + C\)
where \(C\) is the constant of integration.
Finding the Coefficients A and B
We are given that the integral is equal to \(A \sin x + B \cos x + C\).
Comparing our result \(\sin x + \cos x + C\) with \(A \sin x + B \cos x + C\):
- The coefficient of \(\sin x\) is 1. So, \(A = 1\).
- The coefficient of \(\cos x\) is 1. So, \(B = 1\).
Checking the Options
Now we substitute \(A=1\) and \(B=1\) into each of the given options to find the correct one.
| Option | Equation | Substitute A=1, B=1 | Result | Correct? |
|---|---|---|---|---|
| 1 | \(A + B = 0\) | \(1 + 1 = 0\) | \(2 = 0\) | No |
| 2 | \(A + B - 2 = 0\) | \(1 + 1 - 2 = 0\) | \(2 - 2 = 0\) | Yes |
| 3 | \(A + B + 2 = 0\) | \(1 + 1 + 2 = 0\) | \(4 = 0\) | No |
| 4 | \(A + B - 1 = 0\) | \(1 + 1 - 1 = 0\) | \(1 = 0\) | No |
The equation \(A + B - 2 = 0\) is true when \(A=1\) and \(B=1\).
Conclusion on A and B Relation
Based on our calculations, with \(A=1\) and \(B=1\), the correct relationship between A and B is \(A + B - 2 = 0\).
Revision Table: Key Concepts in Integration
| Concept | Description |
|---|---|
| Trigonometric Identities | Fundamental equations relating trigonometric functions, like \(\sin^2 x + \cos^2 x = 1\) and \(\sin 2x = 2 \sin x \cos x\). Essential for simplifying expressions. |
| Perfect Square Trinomial | An algebraic expression that is the square of a binomial, e.g., \(a^2 - 2ab + b^2 = (a-b)^2\). Recognizing these helps simplify expressions under square roots. |
| Square Root of a Square | \(\sqrt{y^2} = |y|\). The result is the absolute value of y. |
| Absolute Value | \(|y|\) is y if \(y \ge 0\) and -y if \(y < 0\). Understanding the domain or range of variables is crucial to remove the absolute value sign correctly. |
| Integration of Basic Trig Functions | Standard integrals like \(\int \cos x \, dx = \sin x + C\) and \(\int \sin x \, dx = -\cos x + C\). |
Additional Information: Integral of \(\sqrt{1 - \sin 2x}\) in Other Ranges
The integral of \(\sqrt{1 - \sin 2x}\) changes depending on the interval of x due to the absolute value \(|\cos x - \sin x|\).
- If \(0 < x < \frac{\pi}{4}\), then \(\cos x > \sin x\), so \(|\cos x - \sin x| = \cos x - \sin x\). The integral is \(\sin x + \cos x + C\).
- If \(\frac{\pi}{4} < x < \frac{5\pi}{4}\), then \(\cos x < \sin x\) (or \(\cos x - \sin x\) is negative). So \(|\cos x - \sin x| = -(\cos x - \sin x) = \sin x - \cos x\). The integral is \(\int (\sin x - \cos x) \, dx = -\cos x - \sin x + C\).
- If \(\frac{5\pi}{4} < x < \frac{9\pi}{4}\), then \(\cos x > \sin x\) again. So \(|\cos x - \sin x| = \cos x - \sin x\). The integral is \(\sin x + \cos x + C\).
This shows that the form of the integral depends on the sign of \((\cos x - \sin x)\), which in turn depends on the specific interval for x.
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